confusion about sample space for independent events

It is the Example 1.10 from the book Introduction to probability Models.

Let a ball drawn from an urn containing four balls, numbered 1,2,3,4. Let E={1,2},F={1,3},G={1,4}. If all four outcomes are assumed equally likely, then

P(EF) = P(E)P(F) = 1/4
P(EG) = P(E)P(G) = 1/4
P(FG) = P(F)P(G) = 1/4


However,

1/4 = P(EFG) not equal to P(E)P(F)P(G)


Hence, even though the events E,F,G are pairwise independent, they are not jointly independent.

My question:

1. Does P(EFG),P(EF),P(E) or P(F) have an intuitive meaning here respectively ? What is sample space for P(EF),P(E),P(EFG)?

I doubt P(EF) means the probability of draw 2 balls from the urn and one of them is {1}, P(EFG) means draw 3 ball from the urn and one of them is {1}?

1. How to calculate P(E) or P(F) here? Is their value equal to 1/2 respectively?

• By $EF$ they mean $E\cap F$, that is the intersection of the events. $E$ occurring and $F$ occurring simultaneously. You still draw only a single ball. In words, you might phrase it $E$ is the event that you draw a small number (small in this case meaning a 1 or a 2) and $F$ is the event that you draw an odd number. $E\cap F$ is the event that you draw a small odd number (i.e. you draw a one). Similarly, $EFG$ here is shorthand for $E\cap F\cap G$ is the event that you draw a small odd number which is one of the outside numbers (i.e. you draw a one). – JMoravitz Aug 26 '16 at 5:08
• As for calculating $Pr(E)$, note that $E=\{1,2\}=\{1\}\cup\{2\}$. We are told that $Pr(\{1\})=Pr(\{2\})=Pr(\{3\})=Pr(\{4\})$ and we know that our sample space is $\Omega=\{1,2,3,4\}$, implying $1= Pr(\Omega)=Pr(\{1\}\cup\{2\}\cup\{3\}\cup \{4\})=Pr(\{1\}) +\dots+Pr(\{4\})=4Pr(\{1\})$ further implying that $Pr(\{1\})=\frac{1}{4}$. So we have $Pr(E)=Pr(\{1\}\cup\{2\})=Pr(\{1\})+Pr(\{2\})-Pr(\{1\}\cap\{2\})=\frac{1}{4}+\frac{1}{4}-0=\frac{2}{4}=\frac{1}{2}$. Similarly for $F$ – JMoravitz Aug 26 '16 at 5:11